Is there an infinite dimensional inner product space without an orthogonal Hamel basis?

Question

I want to know if there exists an (real or complex) vector space $X$ with infinite dimension and an inner product $\langle\cdot,\cdot\rangle$ such that there is no orthogonal Hamel (algebraic) basis of $X$. I do not seek conditions on to the topological aspects of completeness nor an example Schauder Basis or a Maximal Orthogonal System in Hilbert Spaces. I am aware that this is not usually discussed in the scenario of infinite dimensional vector spaces, but this is exactly what I'm curious about.

I know how to prove that every inner product space of Hamel dimension $\aleph_0$ has an orthonormal Hamel basis using the Gram-Schimidt process but this proof does not work for the uncountable case.

I also know that a maximal orthogonal set of non-zero vectors is not always a Hamel basis, thus a simple application of Zorn's Lemma does not seem to solve the problem. Finally, I have seen many other similar questions here but none of them provide an answer for the question posed here.

Answer 1

Let $V$ be an infinitely dimensional separable Hilbert space, and let $\langle., .\rangle$ be its inner product. Equip it with the topology induced by the inner product. Then the inner product is a continuous function from $V\times V$ into $\mathbb K=\mathbb R$ or $\mathbb C$. We will use:

1. Finite dimentional closed subspaces are closed and have empty interior.
2. A Hamel basis of $V$ cannot be countable, or $V$ could be written as a countable union of finite dimensional subspaces, contradicting Baire's Theorem for complete metrizable spaces.
3. Since $V$ is metrizable and separable, $V$ is hereditarely separable.

Now assume by contradiction that $V$ has an orthogonal Hamel basis (or, more generally, an uncountable orthogonal system of nonzero vectors), $B$. By 2. $B$ is separable. Let $S\subseteq B$ be a countable set such that $B\subseteq \overline S$. By 1., there exists $x\in B\setminus S$. Since $V$ is first-countable (since it is metrizable), there exists a sequence $(e_n: n\in \mathbb N)$ of elements of $S$ converging to $x$.

Since every pair of distinct elements of $B$ are orthogonal, $\langle x, e_n\rangle=0$ for every $n\in \mathbb N$. Then by continuity, $0=\langle x, \lim_{n\rightarrow \infty} e_n\rangle=\langle x, x\rangle$, thus $x\in V$, contradicting the fact that $x\neq 0$.

This argument can be adapted for any separable inner product space with uncountable Hamel dimension.

Answer 2

Let $X$ be an infinite dimensional Hilbert space. Assume by contradiction that $\mathcal{B}=\{e_k\}_{k\in K}$ is an orthonormal Hamel basis of $X.$ Fix a sequence $\{e_{k_j}\}_{j=1}^\infty $ of distinct elements of $\mathcal{B}.$ The series $\sum_{j=1}^\infty 2^{-j}e_{k_j}$ is absolutely convergent, thus it represents an element $x_0\in X.$ By assumption $x_0$ is a finite linear combination of the elements in $\mathcal{B}.$ Thus $\langle x_0,e_k\rangle \neq 0$ for at most finitely many $k\in K.$ However $\langle x_0,e_{k_j}\rangle =2^{-j}$ for any $j\in\mathbb{N},$ which leads to a contradiction.

Answer 3

A slightly different argument to see that an infinite-dimensional separable Hilbert space is a counterexample.

Claim: Let $V$ be any separable inner product space (not necessarily complete). Then any orthonormal set $S \subset V$ is at most countable.

Proof: For orthonormal vectors $x,y \in S$, we have $$\|x-y\|^2 = \langle x-y, x-y \rangle = \|x\|^2 + \|y\|^2 - 2 \langle x,y \rangle = 1 + 1 - 0$$ so that $\|x-y\| = \sqrt{2}$. (This is essentially the Pythagorean theorem.) Let $D$ be our countable dense set. For every $x \in S$ we can choose $x' \in D$ with $\|x-x'\| < \sqrt{2}/2$. By the triangle inequality, the elements $x' : x \in S$ are distinct. So the map $x \mapsto x'$ is a 1-1 map of $S$ into $D$, hence $S$ is at most countable.

Couple this with the Baire category argument mentioned in Green Park's answer that in any infinite-dimensional Hilbert space, a Hamel basis is uncountable. We conclude that in an infinite-dimensional separable Hilbert space, there cannot be an orthonormal Hamel basis.